Question: Evaluate the definite integral. $\int^3_{6}\big(9x^2-4x+9\big)\,dx = $
First, use the power rule: $\int^3_{6}\big(9x^2-4x+9\big)\,dx ~=~3x^3-2x^2+9x\Bigg|^3_{6}$ Second, plug in the limits of integration: $[3\cdot3^3-2\cdot3^2+9\cdot3]-[3\cdot6^3-2\cdot6^2+9\cdot6] = 90-630 = -540$. The answer: $\int^3_{6}\big(9x^2-4x+9\big)\,dx~=~ -540$